Question

Please help me with logarithm

Answer 1

Answer:

p+q+2r

Step-by-step explanation:

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Answer 2

Given :

$4 log(3) = p$

$4 log(5) = q$

$4 log(8) = r$

To find :

$4 log(15) + 4 log(64)$

Concept used :

$log(a \times b) = log(a) + log(b)$

$log( {a}^{b} ) = b \: log(a)$

Solution :

$\implies4 log(15) + 4 log(64)$

$\implies4 log(3 \times 5) + 4 log(8 \times 8)$

we know that :-

$log(a \times b) = log(a) + log(b)$

$\implies4 log(3 ) + 4 log(5) + 4 log( {8}^{2} )$

we also know that :-

$log( {a}^{b} ) = b \: log(a)$

$\implies4 log(3 ) + 4 log(5) + (4 \times 2) log( {8})$

$\implies4 log(3 ) + 4 log(5) + 2(4 log( {8}) )$

Now put :-

$4 log(3) = p$

$4 log(5) = q$

$4 log(8) = r$

$\implies \: p + q + 2(r )$

$\implies \: p + q + 2r$

Answer :

$p + q + 2r$